Chapter 10

Question 10.1

Histones have an unusually high concentration of positively charged amino acid residues on their surface compared with most other proteins. Although many SDS molecules bind each protein and give it an overall negative charge, the SDS does not eliminate the positive charges on a protein, it just overwhelms them. For histones, the abundance of positive charges prevents the full effect of SDS on the charge of the proteins, and this manifests as a slower histone migration during electrophoresis compared with most other types of protein.

Question 10.2

Approximately 150 bp are associated with each nucleosome in eukaryotic chromatin, with another 50 bp as linker, bringing the total to ∼200 bp per nucleosome.

Question 10.3

Phosphorylation adds negative charges that can alter the net charge. Acetylation of Lys residues can alter the net charge by creating an amide linkage. Methylation of Lys generally does not remove the positive charge of the terminal amino moiety, although it can alter the pKa of the group.

Question 10.4

Most histone modifications occur near the N-terminus. This part of the histone molecule is a relatively unstructured tail that extends out from the nucleosome core.

Question 10.5

The bacterial chromosome is divided into topologically constrained loops, defined by bound proteins at their boundaries. When the DNA in one loop is relaxed, the DNA in other loops remains supercoiled.

Question 10.6

Transcription will decrease. H2A and H2B are core histones and are closely paired in the nucleosome structure. H1 is generally bound in linker regions, between the core histones, and its level can be varied independent of the core histones. An increase in H1 will lead to greater compaction of the DNA and thus decreased transcription.

Question 10.7

Histone H1 is in the center of the filament, along with the linker DNA. The nucleosomes are stacked along the outside of the filament.

Question 10.8

Bacteria generally divide much more rapidly than eukaryotic cells. Stable protein-rich structures would impede the required replication and segregation of chromosomes at cell division.

Question 10.9

Transcriptionally active genes are characterized by a decrease in histone H1, an absence of bound nucleosomes at the promoter regions, the presence of specialized chromatin remodeling complexes, and the presence of histone variants such as H2AZ and H3.3.

Question 10.10

“Epigenetic inheritance” refers to chromatin modifications (particularly histone modifications) that are retained in the chromatin after cell division and affect gene transcription. Such modifications are not encoded in the DNA and thus are not subject to Mendelian inheritance.

Question 10.11

(c)

Question 10.12

62 × 106 H2A molecules. (“Genome” refers to the haploid genetic content of the cell; the cell is diploid, so the number of nucleosomes is doubled.)

Details: [(3.1 × 109 bp) / (200 bp/nucleosome)] × 2 H2A/nucleosome × 2 [for diploid cell] = 62 × 106 H2A. The 62 million would double on replication.

Question 10.13

Instead of observing eight different complexes, Kornberg would have observed five: H3, H4, H3-H3, H3-H4, and H3-H3-H4.

S-11

Question 10.14

(a) 220 bp is the approximate spacing of adjacent nucleosomes in chromatin. (b) The excess of DNA sequences allowed the investigators to select for sequences that bound tightly and to eliminate weaker binders. (c) The salt interfered with protein-DNA interactions and ensured that only the most tightly binding DNA remained bound to the nucleosomes. (d) Isolation of the DNA-nucleosome complexes reduced the total amount of DNA in each cycle. The PCR step allowed the DNA levels to be increased again. However, only the bound DNA sequences, the “winners,” were amplified; in each cycle, the solution was enriched in DNA sequences binding more tightly to the nucleosomes.